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3 years ago

a class rolled number cubes. their results are shown in the table. what is rhe experimental probability of rolling the number 2

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

3 0

Hi there,
The probability of a dice rolling on a two is 1/6 I think.
If there is two dice, it's 1/36.

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Find the 8th term of the geometric sequence 3, 12, 48,...

Sophie [7]

$\begin{gathered} \text{The first term is 3, the common ratio is 4} \\ \text{That means;} \\ a=3,r=4 \\ \text{The nth term is } \\ T_n=ar^{(n-1)} \\ 8th=3\times4^{(8-1)} \\ 8th=3\times4^7 \\ 8th=3\times16384 \\ 8th=49152 \end{gathered}$

4 0

1 year ago

4 times the measure of the complement of an angle is equal to 2/3 the measure of the original's angles's supplement. what is the

scoundrel [369]

Answer:

72 degrees.

Step-by-step explanation:

If the angle is x degrees its complement is 90-x degrees and it's supplement is 180-x degrees.

4(90-x) = 2/3(180 - x)

360 - 4x = 120 - 2/3 x

360 - 120 = -2/3 x + 4x

240 = -2/3 + 12/3 x

240 = 10/3 x

x = 240 * 3/10

x = 72 degrees.

7 0

3 years ago

Q20-23 with steps pls

just olya [345]

20: Let $x$ be the amount of money. If this amount is shared between 8 people, each person will get x/8 dollars. But there actually are 6 people, so everyone is getting x/6 dollars. We know that this difference results in 30 dollars more, so we have

$\dfrac{x}{6}=\dfrac{x}{8}+30 \iff \dfrac{x}{6}-\dfrac{x}{8}=30$

Rearrange the left hand side as

$\dfrac{4x-3x}{24}=30$

And multiply both sides by 24 to get

$x=720$

21: Let $M$ and $J$ be the number of marbles owned by Mike and Judy, respectively. At the beginning, we have

$M=3J+11$

If they both get 9 more marbles, Mike will have $M+9$ marbles, and Judy will have $J+9$ marbles. So, we have

$M+J+18=93 \iff M+J=75 \iff M=75-J$

Plug this value in the first equation and we have

$75-J=3J+11 \iff 4J=64 \iff J=16$

And we deduce

$M=16\cdot 3 + 11=59$

So, the difference is

$M-J=59-16=43$

22: Let $x,y,z$ be the number of $10, $20 and $50 coupon, respectively. We're given:

$\begin{cases}x=2y+3\\z=\frac{1}{2}y\\x+y+z=38\end{cases}
We can write the third equation by substituting the expressions for x and z:
[tex]x+y+z=(2y+3)+y+\left(\dfrac{1}{2}y\right)=38$